Question: You have found the following ages (in years) of all 5 gorillas at your local zoo: $ 8,\enspace 4,\enspace 14,\enspace 16,\enspace 8$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{8 + 4 + 14 + 16 + 8}{{5}} = {10\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $-2$ years $4$ years $^2$ $4$ years $-6$ years $36$ years $^2$ $14$ years $4$ years $16$ years $^2$ $16$ years $6$ years $36$ years $^2$ $8$ years $-2$ years $4$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{4} + {36} + {16} + {36} + {4}} {{5}} $ $ {\sigma^2} = \dfrac{{96}}{{5}} = {19.2\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{19.2\text{ years}^2}} = {4.4\text{ years}} $ The average gorilla at the zoo is 10 years old. There is a standard deviation of 4.4 years.